What is the integral of (9x^2-8x+16)/(x^3+4x) ?

The integral of (9x^2-8x+16)/(x^3+4x) is 5/2 ln|x^2+4|-4arctan(x/2)+4ln|x|+C

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integral of (9x^2-8x+16)/(x^3+4x)

\int \frac{9 x ^{2} - 8 x + 16}{x ^{3} + 4 x} d x

\frac{5}{2} ln x^{2} + 4 - 4 arctan (\frac{x}{2}) + 4 ln ∣ x ∣ + C

Solution steps

\int \frac{9 x ^{2} - 8 x + 16}{x ^{3} + 4 x} d x

Expand

\frac{9 x ^{2} - 8 x + 16}{x ^{3} + 4 x} : \frac{9 x ^{2}}{x ^{3} + 4 x} - \frac{8 x}{x ^{3} + 4 x} + \frac{16}{x ^{3} + 4 x}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{9 x ^{2}}{x ^{3} + 4 x} d x - \int \frac{8 x}{x ^{3} + 4 x} d x + \int \frac{16}{x ^{3} + 4 x} d x

\int \frac{9 x ^{2}}{x ^{3} + 4 x} d x = \frac{9}{2} ln x^{2} + 4

\int \frac{8 x}{x ^{3} + 4 x} d x = 4 arctan (\frac{x}{2})

\int \frac{16}{x ^{3} + 4 x} d x = 16 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4)

= \frac{9}{2} ln x^{2} + 4 - 4 arctan (\frac{x}{2}) + 16 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4)

Simplify

\frac{9}{2} ln x^{2} + 4 - 4 arctan (\frac{x}{2}) + 16 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4) : \frac{5}{2} ln x^{2} + 4 - 4 arctan (\frac{x}{2}) + 4 ln ∣ x ∣

= \frac{5}{2} ln x^{2} + 4 - 4 arctan (\frac{x}{2}) + 4 ln ∣ x ∣

Add a constant to the solution

= \frac{5}{2} ln x^{2} + 4 - 4 arctan (\frac{x}{2}) + 4 ln ∣ x ∣ + C

\int e^{c o s (11 t)} sin (11 t) d t

\frac{d}{d x} (\frac{1}{9})

\int \frac{a + b x ^{4}}{5 a x + b x ^{5}} d x

\frac{d}{d x} (ln (21 x))

g (x) = (x^{2} + 3) (x^{2} - 4 x)

What is the integral of (9x^2-8x+16)/(x^3+4x) ?
The integral of (9x^2-8x+16)/(x^3+4x) is 5/2 ln|x^2+4|-4arctan(x/2)+4ln|x|+C